Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x-6y &= 4 \\ -9x-8y &= -6\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-9x = 8y-6$ Divide both sides by $-9$ to isolate $x$ $x = {-\dfrac{8}{9}y + \dfrac{2}{3}}$ Substitute this expression for $x$ in the first equation. $6({-\dfrac{8}{9}y + \dfrac{2}{3}}) - 6y = 4$ $-\dfrac{16}{3}y + 4 - 6y = 4$ Simplify by combining terms, then solve for $y$ $-\dfrac{34}{3}y + 4 = 4$ $-\dfrac{34}{3}y = 0$ $y = 0$ Substitute $0$ for $y$ in the top equation. $6x-6( 0) = 4$ $6x = 4$ $6x = 4$ $x = \dfrac{2}{3}$ The solution is $\enspace x = \dfrac{2}{3}, \enspace y = 0$.